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a: \(VT=\dfrac{cot^2x}{1+cot^2x}\cdot\dfrac{1+tan^2x}{tan^2x}\)
\(=\dfrac{cot^2x}{\dfrac{1}{sin^2x}}\cdot\dfrac{\dfrac{1}{cos^2x}}{tan^2x}\)
\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{1}{cos^2x}:\dfrac{1}{sin^2x}\)
\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{sin^2x}{cos^2x}\)
\(=cot^2x\)
\(VP=\dfrac{tan^2x+cot^2x}{1+tan^4x}=\dfrac{\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{sin^2x}}{1+\dfrac{sin^4x}{cos^4x}}\)
\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}:\dfrac{cos^4x+sin^4x}{cos^4x}\)
\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}\cdot\dfrac{cos^4x}{cos^4x+sin^4x}=\dfrac{cos^2x}{sin^2x}=cot^2x\)
=>VT=VP
b:
\(\dfrac{tan^2x-cos^2x}{sin^2x}+\dfrac{cot^2x-sin^2x}{cos^2x}\)
\(=\dfrac{\left(\dfrac{sinx}{cosx}\right)^2-cos^2x}{sin^2x}+\dfrac{\left(\dfrac{cosx}{sinx}\right)^2-sin^2x}{cos^2x}\)
\(=\dfrac{sin^2x-cos^4x}{cos^2x\cdot sin^2x}+\dfrac{cos^2x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x+cos^2x-cos^4x-sin^4x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{1-\left(cos^2x+sin^2x\right)^2+2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}=2\)
\(\frac{2}{sin4x}-tan2x=\frac{2}{2sin2x.cos2x}-\frac{sin2x}{cos2x}=\frac{1}{cos2x}\left(\frac{1}{sin2x}-sin2x\right)\)
\(=\frac{1}{cos2x}\left(\frac{1-sin^22x}{sin2x}\right)=\frac{1}{cos2x}\frac{cos^22x}{sin2x}=\frac{cos2x}{sin2x}=cot2x\)
\(cos5x.cos3x+sin7x.sinx=\frac{1}{2}cos8x+\frac{1}{2}cos2x-\frac{1}{2}cos8x+\frac{1}{2}cos6x\)
\(=\frac{1}{2}\left(cos6x+cos2x\right)=cos4x.cos2x\)
\(\frac{1-2sin^22x}{1-sin4x}=\frac{cos^22x-sin^22x}{cos^22x+sin^22x-2sin2x.cos2x}\)
\(=\frac{\left(cos2x-sin2x\right)\left(cos2x+sin2x\right)}{\left(cos2x-sin2x\right)^2}=\frac{cos2x+sin2x}{cos2x-sin2x}=\frac{\frac{cos2x}{cos2x}+\frac{sin2x}{cos2x}}{\frac{cos2x}{cos2x}-\frac{sin2x}{cos2x}}=\frac{1+tan2x}{1-tan2x}\)
\(2cosx-3cos\left(\pi-x\right)+5sin\left(4\pi-\frac{\pi}{2}-x\right)+cot\left(\pi+\frac{\pi}{2}-x\right)\)
\(=2cosx+3cosx-5sin\left(\frac{\pi}{2}+x\right)+cot\left(\frac{\pi}{2}-x\right)\)
\(=5cosx-5cosx+tanx=tanx\)
\(VT=\frac{2\cos2\alpha.\cos\alpha}{2.\sin2\alpha\cos\alpha}.\frac{\sin2\alpha}{\cos2\alpha}-2\left(2\sin\alpha.\cos\alpha\right)^2\)
\(VT=1-2\left(\sin2\alpha\right)^2=\cos4\alpha\)
a: Ta sẽ có hình vẽ sau:
Đặt \(x=\widehat{B}\)
sin x=sin B=AC/BC
cosx=cosB=AB/BC
\(tanx=tanB=\dfrac{AC}{AB}=\dfrac{sinx}{cosx}\)
=>\(tan^2x=\dfrac{sin^2x}{cos^2x}\)
b: \(cot^2x=\dfrac{1}{tan^2x}=1:\dfrac{sin^2x}{cos^2x}=\dfrac{cos^2x}{sin^2x}\)
\(\frac{1+sin^4x-cos^4x}{1-sin^6x-cos^6x}=\frac{1+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)}{1-\left(sin^2x+cos^2x\right)^2+3sin^2x.cos^2x\left(sin^2x+cos^2x\right)}\)
\(=\frac{1+sin^2x-cos^2x}{1-1+3sin^2x.cos^2x}=\frac{\left(1-cos^2x\right)+sin^2x}{3sin^2x.cos^2x}=\frac{2sin^2x}{3sin^2x.cos^2x}=\frac{2}{3cos^2x}\)
\(\frac{sin^2a-cos^2a}{1+2sina.cosa}=\frac{\left(sina-cosa\right)\left(sina+cosa\right)}{sin^2a+cos^2a+2sina.cosa}=\frac{\left(sina-cosa\right)\left(sina+cosa\right)}{\left(sina+cosa\right)^2}\)
\(=\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}\)
\(\frac{1-cosa+cos2a}{sin2a-sina}=\frac{1-cosa+2cos^2a-1}{2sina.cosa-sina}=\frac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\frac{cosa}{sina}=cota\)
\(VT=tan^4x+cos^4x-2\left(tan^2x+cot^2x\right)+8\)
\(=\left(tan^2x+cot^2x\right)^2-2\left(tan^2x+cot^2x\right)+6\)
\(=\left(tan^2x+cot^2x-1\right)^2+5\)
Mặt khác áp dụng BĐT \(a^2+b^2\ge2ab\Rightarrow tan^2x+cot^2x\ge2\)
\(\Rightarrow\left(tan^2x+cot^2x-1\right)^2+5\ge\left(2-1\right)^2+5=6>5\Rightarrow VT>5\) (1)
Lại có \(3sinx-4cosx=5\left(sinx.\frac{3}{5}-cosx.\frac{4}{5}\right)\)
Do \(\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=1\Rightarrow\) đặt \(\left\{{}\begin{matrix}\frac{3}{5}=cosa\\\frac{4}{5}=sina\end{matrix}\right.\)
\(\Rightarrow VP=3sinx-4cosx=5\left(sinx.cosa-cosx.sina\right)=5sin\left(x-a\right)\)
Do \(sin\left(x-a\right)\le1\Rightarrow5sin\left(x-a\right)\le5\Rightarrow VP\le5\) (2)
(1), (2) \(\Rightarrow VT>VP\)